Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be an analytic function such that $$\lim_{|\vec{x}|\rightarrow \infty}f(\vec{x}) = \infty$$ Let $M$ be the set of points at which $f$ achieves its minimum value. Assume $M$ has more than one element.

1)Assume $M$ is a discrete set. Prove there exists a critical point of $f$ which is not in $M$ or construct a counterexample.

2)Is this still valid if we drop the assumption of $M$ being discrete?

  • $\begingroup$ For (1) and $n=1$ you can use Rolle's theorem between two consecutive elements of $M$. $\endgroup$ – conditionalMethod Nov 17 '19 at 13:39

This is an immediate consequence of the mountain pass lemma in the calculus of variations, detailed e.g. in Evans' book [1] on page 480, and thus holds in very general (possibly infinite dimensional) spaces. However, the proof of that theorem is kind of tricky and there's a number of possible pitfalls on the way.

In our case, I think we can slightly simplify the proof as follows. Let $x,y$ be two points in $M$, and let $k = \inf_\gamma\max_{t\in[0,1]}f\circ \gamma(t)$, where $\gamma$ is a $C^1$-curve from $x$ to $y$. By discreteness of $M$, there exists a circle $C_\epsilon(x)$ around $x$, not containing $y$, with $\min_{\omega \in C_\epsilon(x)}f(\omega) > \min f =: m$. Since $\gamma$ has to cross $C_\epsilon(x)$, we infer $k>m$.

The critical points $\{x \in \mathbb{R}^n: \nabla f(x) = 0\}$ of $f$ form a closed set, since $\nabla f$ is continuous, and from $\lim_{|x|\rightarrow \infty}f(x) = \infty$ we conclude that $f^{-1}([a,b])$ is a compact set for all $a,b\in \mathbb{R}$. In combination this implies that the critical values of $f$ form a closed set as well. Supposing that $k$ is not a critical value, there exists $\delta >0$ such that $[k-\delta,k+\delta]$ contains no critical values. The set $A := f^{-1}([k-\delta,k+\delta])$ is a compact set again, hence there exists $\kappa > 0$ such that $|\nabla f(x)| \geq \kappa$ on $A$. Considering the gradient flow $\Phi(x,t)$ given by $\partial_t \Phi(x,t) = -(\nabla f)\circ\Phi(x,t)$, the above estimate yields $f \circ \Phi(x,\frac{2\delta}{\kappa^2})\leq k-\delta$ for all $x\in A$.

Now, let $\gamma$ be a curve such that $\max_{t\in[0,1]}f\circ \gamma(t) < k + \delta$. Then $\gamma_1(\tau) := \Phi(\gamma(\tau), \frac{2\delta}{\kappa^2})$ defines a curve with $\max_{t\in[0,1]}f\circ \gamma_1(t) \leq k - \delta$, which still connects $x$ and $y$, contradicting the definition of $k$.

The second claim is certainly wrong for smooth functions, consider for example the smooth function given by $f(x) = x e^{-\frac{1}{x^2-1}}$ for $|x|>1$ and $f(x) = 0$ for $x \in [-1,1]$. However, I have a feeling that it might actually still be true for analytic functions satisfying $\lim_{|x|\rightarrow \infty}f(x) = \infty$ and $|M|\geq 2$, but for deeper reasons than the above.

[1] Evans, Lawrence C., Partial differential equations, Graduate Studies in Mathematics. 19. Providence, RI: American Mathematical Society (AMS). xvii, 662 p. (1998). ZBL0902.35002.

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